Question

I am attempting to define a search url using Django with the following structure

www.example.com/afdata/search?device=television&category=news&query=channel&limit=30/

My goal is to extract the values television, news, channel and 30 from the above url and pass it to a view defined as follows

def search(request, device='all', category='single', query='', limit=30):
    return HttpResponse("device=%s, category=%s, query=%s, limit=%d", device, category, query, limit)

afdata is the app and in the url configuration file I defined the url as follows:

from django.conf.urls.defaults import *

urlpatterns = patterns('afdata.views',
    (r'^$', 'index'),
    (r'^search?device=(?P<device>.*)&category=(?P<category>.*)&query=(?P<query>.*)&limit=(?P<limit>d+)/$', 'search')
)

When I run using the above search query in the browser, I get 500 Internal Server Error. index responds fine. Any suggestions on what I may be doing wrong?

Answer 1

i suggest that you change your search url to:

(r'^search', 'search')

then in your view do

def search(request):
    device = request.GET.get('device', 'all')
    category = request.GET.get('category', 'single')
    query = request.GET.get('query', '')
    limit = request.GET.get('limit', 30)

   return HttpResponse("device=%s, category=%s, query=%s, limit=%d", device, category, query, limit)

This is much more robust and changeable and flexible.